Probability that there are at least two consecutive heads Please check my approach for this probability problem

Suppose I have a fair, two-sided coin. If I roll it $5$ times, what is the probability that there are at least two consecutive heads?

Please check my approach:
There are $2^5=32$ possible situations. Now, I want to find the probability that there are no consecutive heads, and subtract that from $1$. 
Now, I can't have $4$ or $5$ heads, because there is a guarantee that there will be at least one set of consecutive heads. 
For $3$ heads, there is only one possibility I can think of, $HTHTH$. 
For $2$ heads, if my first head lands first, then my second head could land third, fourth, or fifth. If heads #1 lands second, them my second could land fourth or fifth, and if the first heads lands third, the second can land fifth, so we have $6$ possibilities. 
For $1$ head, it can land anywhere, so there are $5$ possibilities. 
Computing the probability, we get $6+5+1=12$ possibilities with NO consecutive heads, so the probability that we get no consecutive heads is $\frac{12}{32}=\frac38$, so the probability we DO get consecutive heads is $\boxed{5/8}$.
Am I doing this correctly? If so, is there a better way to do this? If not, what went wrong?
 A: Good job. You almost got it all. You forgot the possibility of getting $0$ heads,  the answer should be $1-\frac{13}{32}=\frac{19}{32}$.
Here's another approach.
Suppose we roll the coin $n$ times, what is the number of configuration such that there are no consecutive heads? let the number of such configuration be $f(n)$.


*

*The configuration can start with a tail, then follow by $n-1$ tosses with no consecutive heads.

*The configuration can also start with a head and then tail, then followed by $n-2$ tosses with no consecutive heads. 
$$f(n)=f(n-1)+f(n-2)$$
Note that  $f(1)=2$, $f(2)=3$.
Hence, we have a shifted Fibonacci sequence. 
$$f(n)=\frac{(1+\sqrt5)^{n+2}-(1-\sqrt5)^{n+2}}{2^{n+2}\cdot \sqrt5}$$
The probability that you are looking for is 
$$1-\frac{f(n)}{2^n}=1-\frac{(1+\sqrt5)^{n+2}-(1-\sqrt5)^{n+2}}{2^{2n+2}\cdot \sqrt5}$$
For your problem, we can quickly look up for the $7$-th Fibonacci number.
$$1,1,2,3,5,8,13.$$
Hence the answer should be $$1-\frac{13}{32}=\frac{19}{32}$$
A: Directly:
$$\begin{align}&P(HH***)+\\
&P(THH**)+\\
&P(TTHH*)+P(HTHH*)+\\
&P(TTTHH)+P(THTHH)+P(HTTHH)=\\
&\frac1{4}+\frac18+\frac2{16}+\frac3{32}=\frac{19}{32}.\\
\end{align}$$
A: As an alternative approach, model the sequence of flips as an absorbing Markov process with transition matrix $$P=\begin{bmatrix}\frac12&\frac12&0\\\frac12&0&\frac12\\0&0&1\end{bmatrix}.$$ The third, absorbing state corresponds to getting two heads in a row. We’re looking for $\Pr(X_5=3 \mid X_0=1)$, which is just the $(1,3)$ entry of $P^5$. It’s not a terrible amount of work to compute $$P^5 = \begin{bmatrix}\frac14&\frac5{32}&\frac{19}{32}\\\frac5{32}&\frac3{32}&\frac34\\0&0&1\end{bmatrix}$$ (the last row never changes, so you can omit quite a few multiplications) which gives the probability as $\frac{19}{32}$.
